Entropy has units of energy-per-temperature, e.g. $X$ joules per Kelvin. Can this be reduced to a simple English sentence that describes the relationship between those $X$ joules and 1 Kelvin?

For example, when we say that an object's velocity is 10 meters per second north, we mean,

At that instant, if you measure a small enough change in the object's position, and divide by the corresponding change in time, the ratio approaches 10 meters north per second.

In every other SI unit I've encountered that was a ratio of one thing to some other thing, it was always possible to write a sentence like that to connect them.

So, if a system's entropy is 10 joules per Kelvin, how would you complete this sentence:

If you _______________ small amount of energy, and _______________ small change in temperature, the ratio approaches 10 joules per Kelvin.

After all, heat capacity is also measured in joules per Kelvin, and that's much easier to understand: "5 joules per Kelvin means if you apply 5 joules of heat energy to the system you raise the temperature by 1 Kelvin." But I can't figure out a similar understanding of joules per Kelvin for entropy.

  • $\begingroup$ Have you checked Clausius Theorem/Inequality construction? It says that for a reservoir of constant temperature $T$ acting on a system, either if you have a reversible or irreversible process in curse, you can build up that several reversible machines acting between two points in the process "carry" changes $\Delta Q_{rev}$ in reversible heat (summing up in a quasi-static process). Entropy comes as the relation between them, reversible heat by absolute temperature in a given reservoir. And the construction leads to show that this quotient is a thermodynamic variable (coordinate). $\endgroup$
    – omivela17
    Jul 4, 2018 at 3:19
  • $\begingroup$ Just to note it, we qualify temperature units like Celsius and Fahrenheit with "degrees" to point out that their zeros aren't at absolute zero. By contrast, temperature units like Kelvin and Rankine have their zeros at absolute zero, such that they're not "degree" units. $\endgroup$
    – Nat
    Jul 8, 2018 at 20:13

1 Answer 1


The derivative encountered in calculus is the limit of the ratio of two distinct changes which are interdependent. For e.g. for a vehicle, distance travelled $x$ is a function of time $t$, and the derivative $dx/dt$ gives its velocity $v$. But suppose you write this relation as $dx/v=dt$. Now we have a strange quantity $dx/v$, equal to change in time $dt$, which cannot be interpreted as "change of something when something else changes". But the strangeness is only apparent; to make it look natural rewrite the relation as $dx=v~dt$ or $dx/dt=v$.

Same goes for change in entropy, $dS=dQ/T$, in which $Q$ is heat energy, $T$ is temperature, and the heat transfer is reversible. If it feels strange then rewrite it as $dQ=T~dS$ and interpret accordingly. You can't interpret entropy change as "change of something when something else changes" simply because of the way it is defined.

  • $\begingroup$ OK but I still don't understand how to interpret $dS=dQ/T$ in terms of what happens. Are you saying that if I add a small quantity of heat energy $dQ$, then the entropy change in the system will be $dS$, where $dS=dQ/T$ ? $\endgroup$
    – Bennett
    Jul 4, 2018 at 7:46
  • $\begingroup$ @Bennett Yes, that's right. It is like saying, if I move a small distance $dx$ at velocity $v$ then the elapsed time will be $dt=dx/v$. $\endgroup$
    – Deep
    Jul 4, 2018 at 7:55

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